Krull dimension of $\mathbb{C}[x,y,z]/I$ where $I=(x^2-yz,xz-x)$.

Question

Krull dimension of $\mathbb{C}[x,y,z]/I$ where $I=(x^2-yz,xz-x)$.

The problem says first verify $p_1=(x,y)$, $p_2=(x,z)$ and $p_3=(x^2-y,z-1)$ are prime minimal over $I$. How can I use it ?

Answer 1

Here is an elementary proof which only uses the definition of Krull dimension.

Let $k$ be a field. A prime ideal in $k[x,y,z]/(x^2-yz,xz-x)$ corresponds to a prime ideal of $k[x,y,z]$ which contains $x^2-yz$ and $xz-x=x(z-1)$. There are two cases: a) It contains $x$. Then it corresponds to a prime ideal in $\mathbb{C}[y,z]/(yz)$, which has Krull dimension $1$. b) It contains $z-1$. Then it corresponds to a prime ideal in $k[x,y]/(x^2-y) \cong k[x]$, which has Krull dimension $1$. It follows that $k[x,y,z]/(x^2-yz,xz-x)$ has Krull dimension $1$.

You can do the same proof geometrically: $$V(x^2-yz,xz-x)=V(x^2-yz) \cap V(x(z-1))=(V(x^2-yz) \cap V(x)) \cup (V(x^2-yz) \cap V(z-1))=V(x,yz) \cup V(x^2-y,z-1)$$ Now finish as above.